Limits
Determine \lim_{x \to \infty} \left(\frac{x+3}{x-1}\right)^{x+2}, a classic exponential indeterminate form of the type 1^{\infty}.
Select the correct option:
Solution
e4
An expression of the form 1^{\infty} is handled by the standard identity \lim (1+f)^{g} = e^{\lim f \cdot g} whenever f \to 0 and g \to \infty. Write the base as 1 + \frac{4}{x-1}, so f = 4/(x-1) and the exponent is g = x+2. Then the exponent of e is \lim_{x \to \infty} \frac{4(x+2)}{x-1} = 4, because the ratio of leading coefficients tends to 1 and the constant offsets vanish in the limit. Hence the value is e^4. Option e^2 incorrectly uses a difference of 2 between numerator and denominator constants rather than the gap of 4. Option e^3 mismatches the additive constants and drops a factor. Option e^{-4} reverses the sign by writing the base as 1 - 4/(x+3), which is algebraically incorrect since the fraction added is positive. The decisive JEE Advanced pattern is reducing a 1^{\infty} limit to the exponential of the product of the small part and the large exponent. Plausibility check: the base exceeds 1 for large x and the exponent grows without bound, so the result must exceed 1, ruling out e^{-4} immediately and confirming a growing power of e.
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About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- limits
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
e4
An expression of the form 1^{\infty} is handled by the standard identity \lim (1+f)^{g} = e^{\lim f \cdot g} whenever f \to 0 and g \to \infty. Write the base as 1 + \frac{4}{x-1}, so f = 4/(x-1) and the exponent is g = x+2. Then the exponent of e is \lim_{x \to \infty} \frac{4(x+2)}{x-1} = 4, because the ratio of leading coefficients tends to 1 and the constant offsets vanish in the limit. Hence the value is e^4. Option e^2 incorrectly uses a difference of 2 between numerator and denominator constants rather than the gap of 4. Option e^3 mismatches the additive constants and drops a factor. Option e^{-4} reverses the sign by writing the base as 1 - 4/(x+3), which is algebraically incorrect since the fraction added is positive. The decisive JEE Advanced pattern is reducing a 1^{\infty} limit to the exponential of the product of the small part and the large exponent. Plausibility check: the base exceeds 1 for large x and the exponent grows without bound, so the result must exceed 1, ruling out e^{-4} immediately and confirming a growing power of e.
This medium difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of limits. It appeared in the 2025 exam.
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